Given a z-shaped telescope setup with two beamsplitters such as depicted in Fig. 1, we calculate the effects on the outgoing beam when applying a small additional tilt β to the angle of incidence α of both beamsplitters.
Such a small extra tilt is applied in Finesse 2 via the xbeta parameter of a bs component. With respect to the “untilted beam” (i.e. β = 0) we calculate – for the outgoing beam – both the shift in its position ∆ and the change of its phase δφ, each as a function of the angle β and the distance s between the beamsplitters. For the shift we derive eq. (5) below
while for the phase change we find eq. (13) below
Note that this is the first post in a series, the second post (part 2: finesse examples) will compare the results obtained here with simulations from finesse 2.
Fig. 2 depicts the two different trajectories. The distance between the two beamsplitters is s. The initial angle of incidence of the “untilted” beamsplitters (dark bue) is α and its corresponding trajectory is plotted in dark red. The two beamsplitters are rotated over a small additional angle β (light blue) with corresponding trajectory in light red.
As a result, the path traveled by the beam between the first and second beamsplitter changes from s to s̃; it will hit the second mirror at a different point (a distance d away from the original point) and the outgoing beam will shift over a distance Δ. The difference in total path length between the dark and light red paths will cause a phase shift δφ. In the next subsections we will derive Δ and followed by δφ.
Calculation of shift in position Δ
Using the sine rule for the triangle enclosed by d, s and s̃
we can get the length d
Furthermore from the total angle of that same triangle we can get the angle γ
We need to calculate Δ:
where we used eq. (3). Using eq. (2) we then obtain the shift
We see that all α dependence cancels out. Furthermore, we see that the shift increases linearly with the distance between the two beamsplitters. Since for larger values of β the paraxial approximation requires more and more higher modes to be included in a simulation, increasing s instead of β could help simulating the same shift Δ (we will look into convergence in the second post of this series).
Calculation of difference in path length and δφ
Note that there are two parts:
- p1: the difference between s̃ and s
- p2: horizontal part since we hit the beamsplitter further to the right
p1 will have a positive contribution, while p2 will have a negative contribution to the change in path length.
First note that
We can obtain s̃ from the sine rule eq. (1) and use (6) to get
Since part p2 shortens the total path, we add an overall minus sign:
Using eq. (2) and (6) this becomes
combining the two
Combining eq. (8) and (10) we get
We work out the numerator:
Putting this into eq. (11) we see that the cos(α+β) factor cancels out and we find the total path difference
Note that again the α dependence cancels out. The phase shift resulting from the path difference can be calculated by dividing eq. (12) by the wavelength λ and multiplying by 360°. We also pick up an extra overall minus sign, cancelling the one from eq. (12), since a plane wave is defined as cos(ω t – k x + φ), see for example Bond et al. (2016) eq. (1.4). Putting it together we get
- Both the shift of the beam eq. (5) and the change in phase eq. (13) are independent of the angle α
- Given the square in eq. (12) the path length difference is small but not per se compared to the wavelength λ, leading to a significant phase shift.
Effect finite size beam
The above calculation was done for an infinitesimally narrow beam hitting the beamsplitter exactly in the midpoint (its rotation point). We show here that the result is the same for a beam hitting the beamsplitter off-axis.
From Fig. 3 we see that – compared to the on-axis case – the effect of such an off-axis incoming beam can be described by a horizontal translation of the original beamsplitters (dark blue) with their beams (dark red) with respect to the original tilted beamsplitters (light blue) and their beams (light red). It is clear that
- this does not affect the shift ∆ of the beam on the beam detector.
- this does not affect the total path length difference and hence δφ (the dark red beam gains a bit on the left and loses the same amount on the right).